Questions tagged [summation]

Questions about evaluating summations, especially finite summations. For infinite series, please consider the (sequences-and-series) tag instead.

The notation $\sum\limits_{i=1}^na_i$ means $a_1+\ldots +a_n$.

Use for sums of infinite series and questions of convergence; use for questions about finite sums and simplification of expressions involving sums.

17770 questions
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Let $x_i$ be positive number satisfying $\sum x_i = 1$, what is $\sum ix_i$

Let $x_i$ be a positive number for each $i \in \{1, 2, 3 \dots \}$ such that $\sum_{i=1}^\infty x_i = 1$ is there a closed formula for $$\sum_{i = 1}^\infty ix_i$$?
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Is ∑ with negative value solvable?

Is it possible to have a negative value in sigma? e.g. $y = \Sigma_{k=0}^{k=-2} k \times 10$ Will this give the result $(0 \times 10) + (-1 \times 10) + (-2 \times 10) = -30 $? Or will it be $\infty$ because $k$ will be increased with $1$ until…
user94342
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Dervation of $\sum_{k=0}^n(r+1)^k= \frac{(r+1)^{n+1}-1}{r}$

How can one derive the following identity? $\sum_{k=0}^n(r+1)^k= \frac{(r+1)^{n+1}-1}{r}$ I have playing around with binomial coefficients and index shiftings but wasn't able to get anywhere.
Phil-ZXX
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Finding the value of $ \sum_{n=5}^{204} (n - 2) $

Is there a generalized formula for finding a sum such as this one? I'm going over an old quiz for a programming class but I'm not able to solve it: $$ \sum_{n=5}^{204} (n - 2) $$ I know this is probably dead simple, but I'm seriously lacking on the…
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Hint on power sum coefficients

Please do not give anything more than a tiny hint for this question. I know that there is a well-known formula for $$\sum_{i=1}^n i^k,$$ where $k$ is any non-negative integer. I have been able to prove that in fact it is a polynomial in…
dfeuer
  • 9,069
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Summation problem with inequalities

I can't go to second step from the first. Can you please explain the way? I will be obliged if you also explain how to go to third step. Thank you.
Silent
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Summation (telescoping series)

Find the sum of the following series: $$2 \cdot 1! + 5 \cdot 2! + 10 \cdot 3! + 17 \cdot 4! + \cdots + (n^2 +1)n!$$ Since the question is asking about the closed form of its sum, I thought it must be some telescoping series. So I tried to express it…
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How can I find the sum $\sum_{i=1}^{10} i(i+1)$?

During the resolution of a physics problem, I have encountered the sum: $$\sum_{i=1}^{10} i(i+1)$$ For practical purposes, I evaluated it using a Python script, but a calculator would have been enough since the sum stops at $i=10$. However, I wonder…
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Double Summation with a constant as the Summand

As part of the preparation for an Exam, I'm working through a simple double sum exercise that has a constant as the Summand. The exercise has the answer (result) available but not the resolution. I attempted to solve it, but apparently I've reached…
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How to prove that the above summation is equal to $(\pi/101)\tan(\pi/202)$

$$\sum_{n=0}^\infty\left(\frac1{50+101n}-\frac1{51+101n}\right)$$ How to prove that the value of the above summation is equal to $(\pi/101)\tan(\pi/202)?$ I am trying this question by putting n=0,1,2,3,...and so on and getting the series to be…
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Sum of all absolute values of difference of the numbers 1,2,.....n taken two at a time

The sum of all absolute values of the differences of the numbers $1,2,3,\ldots, n$, taken two at a time, i.e. $$\sum_{j\le i\le n}\left|i-j\right|$$ equals: (A) $\binom{n-1}3$ (B) $\binom n 3$ (C) $\binom{n+1}3$ (D)$\binom{n+2}3$ Please give me…
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finding sum of infinity power series

What is the sum of the below series? $$ \sum_{n=1}^\infty \frac{3^{n-1}-1}{6^{n-1}} $$ here $$ r=\frac{1}{2} $$ from what I know about summations, when |r|<1, the sum is equal to: $$ s=\frac{a}{1-r} $$ where a is the first member of the series. Here…
Amirhosein
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How to write a sigma summation with fractional stop

If I had a summation such as: $$\sum_{i=0}^\frac{n}{2} n+1 $$ Is there a way I could do it, or is it just undefined notation? I thought a switch with the fractions like this would work: $$\sum_{i=0}^n \frac{n+1}{2} $$ As in, double the terms but…
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Evaluate $\sum_{n=1}^\infty\frac{n}{(n-1)!}$.

I realise this is quite simple, but I just don't exactly understand what the last step is. $$\sum_{n=1}^\infty\frac{n}{(n-1)!}=\sum_{n=1}^\infty\left(\frac{1}{(n-2)!}+\frac{1}{(n-1)!}\right)=\sum_{n=1}^\infty\frac{1}{(n-2)!}+e$$
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How do I evaluate the summation symbol "sigma" when the index of summation (i) doesn't appear in the indexed variable?

$\sum_{i=1}^{300} {1/3}$ My probability text wrote the above. I don't know how to interpret it since "i" doesn't show up in the indexed variable. How is this interpreted? Do I just add up 300 1/3's? $\sum_{i=1}^{300} {n}$ Does this mean 300n?